Appendix of Real-time quantitative PCR training course.
Division of Genomics Research
Life Science Research Center, Gifu University
This document explains how you can interpret the Ct values from DDCt method and how you can determine the DDCt method is applicable for your Experimental instead of the
conventional calibration curve method.
If you are not sure which method to use, use calibration curve method because it is more reliable for most of the real-time
PCR applications.
Ct value interpretation
Method 1
Ct value interpretation
Experimentalal condition
• Sample subject: Control (Cont) and Experimental (Exp)
• Target gene: Tar
• Reference gene: Ref
Results
• Control sample
Ct for Tar (ContCtTar) was 22, Ct for Ref (ContCtRef) was 19.
• Experimental sample
Ct for Tar (ExpCtTar) was 25, Ct for Ref (ExpCtRef) was 18.
Ct value interpretation
DCt
Tar= (2^ExpCt
Tar/2^ContCt
Tar)
= (2
25/ 2
22) = (2
25-22) = 2
3This calculation assumes that the Tar DNA becomes 2-fold
more after each PCR cycle (when 100% PCR efficiency) so that the expression difference is 23 = 8-fold.
Experimental had 8-fold less Tar DNA compared to that of Cont, because high Ct value means less DNA at the starting point.
DCt
Ref= (2^ExpCt
Ref/2^ContCt
Ref) = (2
18/ 2
19) = (2
18-19) = 2
-1In the same manner the expression difference is 2-1 = 1/2-fold, meaning Exp had 1/2 times less (meaning 2-fold more) Ref
DNA compared to that of Cont at the starting point.
This is also assuming that Ref has 100% PCR efficiency.
Ct value interpretation
DDCt = (DCt
Tar/DCt
Ref) = (2
3/ 2
-1) = 2
3-(-1)= 2
4= 16
This is compensating the amount of Tar using the amount of Ref gene.
Real expression difference is 16-fold.
This means the amount of Tar in Exp was 16-fold less than that in Cont so that if you set the amount of Tar in Cont as 1.0, the amount of Tar in Exp is 0.0625 (= 1/16).
Attention!!:
This calculation is applicable only when the PCR
efficiency of both Tar and Ref is 100% (DNA product becomes 2-fold after each PCR cycle).
Ct value interpretation
If the PCR efficiency is less than 100%, you have to change the calculations.
• Hypothetical results:
The efficiency was 95% for Tar and 93% for Ref, meaning that DNA becomes 1.95-fold more for Tar and 1.93-fold more for Ref after each PCR cycle.
Ct value interpretation
The calculations change as follows:
DCtTar = (1.95^ExpCtTar / 1.95^ContCtTar ) = (1.9525 / 1.9522 ) = (1.9525-22) = 1.953
DCtRef = (1.93^ExpCtRef / 1.93^ContCtRef ) = (1.9318 / 1.9319 ) = (1.9318-19) = 1.93-1
DDCt = (DCtTar /DCtRef) = (1.953 / 1.93-1 ) ≒ (7.41 / 0.52) ≒ 14.31
This means that the amount of Tar in Exp was 14.31-fold less than that in Cont so that if you set the amount of Tar in Cont as 1.0, the amount of Tar in Exp is
approximately 0.070 (= 1/14.31)
Ct value interpretation
Important rules:
You can use the DDCt method only when the PCR efficiency of Tar and Ref is steady, high and almost equal.
How to evaluate the PCR efficiency?
• When you make calibration curves for Tar and Ref, you will have values of the steadiness (R2), the amplification efficiency and the slope for each calibration curve (refer to the next page).
• If the efficiency is 90% or higher and R2 is 0.95 or higher (0.99 or higher is preferred) for both Tar and Ref and at the same time the difference of slope values between Tar and Ref is less than 0.1 (means 0.999 is OK but 0.100 is not OK), the Experimentalal condition is good enough to use the DDCt method, otherwise you have to use the calibration curve method.
Ct value interpretation
Slope R2 Efficiency
